2.1 Complex function

1. For $z = x + iy$ , $w = u + iv$ , calculate $w = z^2$ . Then

$\displaystyle w = z^2 = (x + iy)^2 = x^2 - y^2 + i(2xy) = u + iv$

Thus,

. Next we solve this for $x,y$

Since $v = 2xy$ , $y = \frac{v}{2x}$ . Put this into $u = x^2 - y^2$ . Then

$\displaystyle u = x^2 - (\frac{v}{2x})^2 = x^2 - \frac{v^2}{4x^2}$

Multiply both sides by $4x^2$

. Then

$\displaystyle 4x^4 - 4x^2 u - v^2 = 0$

This can be thought of quadratic equation in $x^2$

. Thus let $x^2 = X$

. Then. by the formula of quadratic equation, we have

$\displaystyle x^2 = \frac{2u \pm \sqrt{4u^2 + 4v^2}}{4} = \frac{u \pm \sqrt{u^2 + v^2}}{2}$

Note that

is a real part. Thne

$\displaystyle x^2 = \frac{u + \sqrt{u^2 + v^2}}{2}$

Thus we have

$\displaystyle x = \pm \sqrt{\frac{u + \sqrt{u^2 + v^2}}{2}}$

Next find

. Since

, we have

$\displaystyle y^{2} = x^2 - u = \frac{u + \sqrt{u^2 + v^2}}{2} - u = \frac{-u + \sqrt{u^2 + v^2}}{2}$

Thus

$\displaystyle y = \pm \sqrt{\frac{-u + \sqrt{u^2 + v^2}}{2}}$

Next, consider what kind of curve the straight line $y = \pm b (b> 0)$ parallel to the real axis of the $z$ plane is mapped. By the above equation, $y = \pm b$ satisfies

$\displaystyle y = \pm b = \pm \sqrt{\frac{-u + \sqrt{u^2 + v^2}}{2}}$

Thus,

$\displaystyle b^2 = \frac{-u + \sqrt{u^2 + v^2}}{2}$

and it is mapped to the parabola $v^2 = 4b^2(b^2 + u)$

Similarly, a straight line parallel to the imaginary axis of the $z$ plane $x = \pm a (a > 0)$ satifies

$\displaystyle x = \pm a = \pm \sqrt{\frac{u + \sqrt{u^2 + v^2}}{2}}$

Thus

$\displaystyle a^2 = \frac{u + \sqrt{u^2 + v^2}}{2}$

and it is mapped to parabola $v^2 = 4a^2(a^2 - u)$

(a) $z = x + iy$ , $w = u + iv$ 縺ｨ縺翫￥縺ｨ

$\displaystyle u + iv = z^3 = (x + iy)^3 = x^3 - 3xy^2 + i(3x^2 y - y^3)$

繧医ｊ

(b) Let $z = x + iy$ , $w = u + iv$ . Then

$\displaystyle u + iv = \frac{z}{z+1} = \frac{z(\bar{z} + 1)}{\vert z + 1\vert^2} = \frac{\vert z\vert^2 + z}{\vert z+1\vert^2}$

Note that $\vert z\vert^2 = x^2 + y^2$ .

$\displaystyle u + iv = \frac{x^2 + y^2 + x + iy}{(x+1)^2 + y^2}$

Therefore, $u = \frac{x^2 + y^2 + x}{(x+1)^2 + y^2}, v = \frac{y}{(x+1)^2 + y^2}$ .

$\displaystyle u + iv = \frac{z - i}{z + i} = \frac{(z-i)(z-i)}{\vert z + i\vert^2} = \frac{(x + i(y-1))^2}{\vert z+i\vert^2}$

Note that $\vert z+i\vert^2 = x^2 + (y+1)^2$ . Then

$\displaystyle u + iv = \frac{x^2 - (y-1)^2 + 2ix(y-1)}{x^2 + (y+1)^2}$

Therefore, $u = \frac{x^2 - (y-1)^2}{x^2 + (y+1)^2}, v = \frac{2x(y-1)}{x^2 + (y+1)^2}$ .